Answer to **Puzzle 196**: Let's call our monoidal poset \\(\mathcal{P}\\). For every \\(a\in\mathcal{P}\\) such that \\(a\leq x\multimap y\\), \\(a\leq x\multimap y\Leftrightarrow x\otimes a\leq y\\). If \\(a,b\in\mathcal{P}\\) and \\(x\otimes a\leq y\\) and \\(x\otimes b\leq y\\), then \\(a\leq x\multimap y\\) and \\(b\leq x\multimap y\\), so \\(x\multimap y\\) is an upper bound in \\(\mathcal{P}\\) for \\(a\\) and \\(b\\). By the same argument, \\(x\multimap y\\) must be an upper bound for \\(\\{a\in\mathcal{P}:x\otimes a\leq y\\}\\), i.e., \\[\bigvee \lbrace a : \; x \otimes a \le y \rbrace\leq x\multimap y\\] . Since \\(x\multimap y\in\mathcal{P}\\), and \\(x\multimap y\leq x\multimap y\\) so that \\(x\otimes(x\multimap y)\leq y\\), \\(x\multimap y\\) is an element of the set of which it is an upper bound. Therefore it is the least upper bound, i.e., \\[x\multimap y=\bigvee \lbrace a : \; x \otimes a \le y \rbrace\\] .

Answer to **Puzzle 194**: \\(S\multimap T=\bigvee \lbrace A : \; S \otimes A \le T \rbrace=\bigvee \lbrace A : \; S \cup A \subseteq T\rbrace\\), so this is the set with the most elements such that its union with \\(S\\) is \\(T\\). This set is \\(S^c\cap T\\) (where \\(S^c\\) denotes the set complement operation), since the addition of any more elements of \\(X\\) would make its union with \\(S\\) a proper superset of \\(T\\).